Physics Q38 – JEE-MAIN 2026

Home Physics Current ElectricityQuestion #37

Two resistors of

200 Ω

and

400 Ω

are connected in series with a battery of

100 V

. A bulb rated at

200 V, 100 W

is connected across the

400 Ω

resistance. The potential drop across the bulb is _______ V.

Clue 1 of 3: Circuit Analysis

First, determine the resistance of the bulb from its rating. Then, identify the series and parallel combinations in the circuit.

🧠 Full Solution Path

Step 1: Calculate the resistance of the bulb✦ Active

The resistance of the bulb can be determined from its rated power and voltage using the formula $P = V^{2} / R$ .

R_{b u l b} = \frac{V_{r a t e d}^{2}}{P_{r a t e d}} = \frac{(200 V)^{2}}{100 W} = \frac{40000}{100} = 400 Ω

Step 2: Calculate the equivalent resistance of the parallel combination○ Expand

The $400 Ω$ resistor is connected in parallel with the bulb, which also has a resistance of $400 Ω$ . The equivalent resistance of this parallel combination is:

R_{p a r a l l e l} = \frac{R_{2} \times R_{b u l b}}{R_{2} + R_{b u l b}} = \frac{400 Ω \times 400 Ω}{400 Ω + 400 Ω} = \frac{160000}{800} = 200 Ω

Step 3: Calculate the total current and potential drop across the bulb○ Expand

The $200 Ω$ resistor ( $R_{1}$ ) is in series with the parallel combination ( $R_{p a r a l l e l}$ ). The total resistance of the circuit is:

R_{t o t a l} = R_{1} + R_{p a r a l l e l} = 200 Ω + 200 Ω = 400 Ω

The total current drawn from the battery is:

I_{t o t a l} = \frac{V_{b a t t e r y}}{R_{t o t a l}} = \frac{100 V}{400 Ω} = 0.25 A

The potential drop across the bulb is the potential drop across the parallel combination:

V_{b u l b} = I_{t o t a l} \times R_{p a r a l l e l} = 0.25 A \times 200 Ω = 50 V

💡 Teacher's Secret Hint

Remember that components in parallel have the same potential difference across them.

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Physics Question 38 – JEE-MAIN 2026

Clue 1 of 3: Circuit Analysis

🧠 Full Solution Path